HINT:
If the Jacobian matrix has negative determinant then it is orientation reversing. If it has positive determinant then it is orientation preserving.
The Jacobian matrix, in this case, is the two-by-two matrix whose columns are $F_{r}$ and $F_{\theta}$. Can you find the partial derivatives, put them in a matrix and find its determinant?
You may find that the Jacobian is singular at some point.
The resulting object will be (assuming everything's compact, say...) an orientable manifold, but not an oriented one. If you want it to be oriented, with the orientation matching that of $M_1$, say, then the map $f$ should be orientation reversing (which may not be possible without reversing the orientation on $M_2$, and hence the induced orientation on $\partial M_2$).
If you think hard about two unit disks in the plane, joined to make a sphere, you'll see what I mean.
post-comment edits
A manifold $M$ is a topological space $X$ together with a (maximal) atlas $A$ that satisfies certain rules that are well-known to the questioner, so I won't write them out. An oriented manifold is the same thing, together with a subset $S$ of the atlas that consists of charts that are declared to be "orientation preserving" (and which, when they have overlapping domains, satisfy something that's the equivalent (in the smooth case) of the transition function's derivative having positive determinant on the overlap).
To any oriented manifold, $(X, A, S)$ there's an associated manifold $(X, A)$ gotten by ignoring the orientation.
If we glue oriented manifolds $M_1$ and $M_2$ together as in the question and if their boundaries are each connected and $f$ is an orientation-reversing homeomorphism between them, then $M = M_1 \cup_f M_2$ is orientable, and indeed, the orientation can be made consistent with that of $M_1$ and $M_2$, in the sense that both $M_1$ and $M_2$ are embedded in $M$, and the embeddings are orientation-preserving. The glued up manifold is orientable, but until we pick an orientation, it's just a manifold! And the homeomorphism class of that manifold is independent of the orientations of $M_1$ and $M_2$. That's the claim that Michael Albanese, in the comments, seems willing to believe. Let's look at the orientation preserving case now.
If $f$ happens to be orientation-preserving, then we can reverse the orientation on $M_2$ to get a different oriented manifold, $M_2'$, and a map $f' : \partial M_1 \to \partial M_2': x \mapsto f(x)$ which is exactly the same as $f$, except that the target oriented-manifold is now $M_2$ with the other orientation. Note, too, that $M_2'$ and $M_2$ are homeomorphic as manifolds, by the identity map, but are not necessarily homeomorphic by an orientation-preserving homeomorphism.
If we now build $M' = M_1 \cup_{f'} M_2'$, it's homeomorphic to $M$, but not necessarily oriented-homeomorphic. It's orientable, and with the right orientation, we find that $M_1$ and $M_2'$ are both embedded in it by orientation-preserving embeddings.
On the other hand, it's not true that $M_2$ is embedded in it by an orientation-preserving embedding.
What IS true is that $M_2$ is embedded in it as a topological submanifold-with-boundary.
In the case Michael asked about, where $\partial M_1 = \partial M_2 = S^3$, and $M_1$ and $M_2$ are both $\Bbb CP^2$, we get that
$$
M' = \Bbb CP^2 \# \overline{\Bbb CP^2}
$$
where the "equality" here is an orientation-preserving homeomorphism (basically just "inclusion"). But it's also true that $M'$ is homeomorphic (again by inclusion) to
$\Bbb CP^2 \# \Bbb CP^2$, by a non-orientation-preserving homeomorphism.
Best Answer
Derivative of a smooth map $F$ at a point $x$ can be seen as a linear map from the tangent space $T_x\Bbb R$ to the tangent space $T_{f(x)}\Bbb R$. You may view both $T_x\Bbb R$ and $T_{f(x)}\Bbb R$ as the same as the vector space $\Bbb R$ itself, and the linear map would be defined by $v\mapsto F'(x)v$. Whether or not this linear map is orientation-preserving or -reversing just depends on whether $F'(x)$ is positive or negative in this one-dimensional case.
$\mathbb{R}^m - \boldsymbol{0}$ can be seen as the union of many lines radiating from $\boldsymbol{0}$, each line is of the form $\{cv:v\text{ is a fixed unit vector}, c\in(0,\infty)\}$. You may see that each of such line looks exactly like $(0,\infty)$, just pointing in some directions other than along x-axis. The function $\alpha_m$ flips each of these lines the same way $\alpha$ flips $(0,\infty)$. The far points are mapped close to $\boldsymbol{0}$, points close to $\boldsymbol0$ are mapped to far away, and all points on a line $\{cv:v\text{ is a fixed unit vector}, c\in(0,\infty)\}$ remain on the same line after the map $\alpha_m$.
Sorry but I don't understand your question. If you have two oriented manifolds $M_1,M_2$ of the same dimension, you may try to find a smooth map $F:M_1\to M_2$ whose derivative's determinant is negative. But if you are talking about reversing the orientation of just one oriented manifold $M$, you should be looking for a diffeomorphism $F:M\to M$, not function from some other manifold like $\Bbb R^n$.